Answer:
Final temperature = 83.1 °C
Explanation:
Given data:
Mass of concrete = 25 g
Specific heat capacity = 0.210 cal/g. °C
Initial temperature = 25°C
Calories gain = 305 cal
Final temperature = ?
Solution:
Q = m. c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
305 cal = 25 g ×0.210 cal/g.°C × T2 - 25°C
305 cal = 5.25cal/°C × T2 - 25°C
305 cal / 5.25cal/°C = T2 - 25°C
58.1 °C = T2 - 25°C
T2 = 58.1 °C + 25°C
T2 = 83.1 °C
Answer:
Q = 4.056 J
Explanation:
∴ m = 406.0 mg = 0.406 g
∴ <em>C </em>= 1.85 J/g.K
∴ T1 = 33.5°C ≅ 306.5 K
∴ T2 = 38.9°C = 311.9 K
⇒ ΔT = 311.9 - 306.5 = 5.4 K
⇒ Q = (0.406 g)(1.85 J/gK)(5.4 K)
⇒ Q = 4.056 J
Answer:
Macromolecules. A very large organic molecule composed of many smaller molecules, 1)Carbohydrates, 2)proteins, 3)lipids, 4)nucleic acids. Three of the four classes of macromolecules that are polymers. 1.Carbohydrates.
